Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-1 23
From (i) and (ii), we have
H = < a d >
But
O(H) = O( < a d > )
O ( < a d > ) = d
n
(Θ (a d ) d
n = e)
Hence
O(H) =
n
gcd(m,n)
which completes the proof of the theorem.
,
Theorem 13. Any two cyclic groups of the same order are isomorphic.
Proof. Let G and H be two cyclic groups of the same order. Consider the mapping
f : G → H
defined by
f(a r ) = b r
Then f is clearly an homomorphism. Also,
f(a r ) = f(a s ) b r = b s ,
If G and H are of infinite order, then
r = s
and so a r = a s .
If their order is finite, say n, then
B r = b s b r−s = e
n | (r−s)
nu = r−s , u ∈ Z
a r−s = a nu
a r = a s .
= (a n ) u = e
Hence f is 1−1 mapping also. Therefore, G ~ H.
Theorem 14. Every isomorphic image of a cyclic group is again cyclic.
Proof. Let G = < a > be a cyclic group and let H be its image under isomorphism f. The elements of G
are given by
G = {…, a −r ,…, a −3 , a −2 , a −1 , a, a 2 , a 3 ,…, a r ,…}
Let be an arbitrary element of H. Since H is isomorphic image of G, there exists a r ∈ G, r = 0, 1,….
Such that b = f(a r ). Since f is homomorphism, we have
b =
f (a).f (b)....f (c)
1 4 4 2 4 4 3
r factors
= (f(a) r
Thus H is generated by f(a) and hence is cyclic.